Problem: $g(t) = -t^{2}-t$ $h(x) = 3x+5+4(g(x))$ $f(x) = -7x^{2}-5x+h(x)$ $ g(f(1)) = {?} $
Explanation: First, let's solve for the value of the inner function, $f(1)$ . Then we'll know what to plug into the outer function. $f(1) = -7(1^{2})+(-5)(1)+h(1)$ To solve for the value of $f$ , we need to solve for the value of $h(1)$ $h(1) = (3)(1)+5+4(g(1))$ To solve for the value of $h$ , we need to solve for the value of $g(1)$ $g(1) = -1^{2}-1$ $g(1) = -2$ That means $h(1) = (3)(1)+5+(4)(-2)$ $h(1) = 0$ That means $f(1) = -7(1^{2})+(-5)(1)$ $f(1) = -12$ Now we know that $f(1) = -12$ . Let's solve for $g(f(1))$ , which is $g(-12)$ $g(-12) = -(-12)^{2}-(-12)$ $g(-12) = -132$